$\begingroup$What are the mixing weights? In other words, what is your prior belief that the draw will come from distribution 1 versus distribution 2?$\endgroup$
– AdrianMar 2 '16 at 21:34

$\begingroup$I'm not sure. How does that affect the analysis? I guess I'd like to see the case where your prior is that I will pick from $X_2$ with probability 1, and the case where your prior is that I'll pick from either with equal probability (i.e., .5).$\endgroup$
– sundanceMar 2 '16 at 21:47

2

$\begingroup$The mixing weight would alter the prior. Using 0.5 implies you're as likely to have a point from either. You can't tell if that's reasonable from the information in your question$\endgroup$
– Glen_b♦Mar 3 '16 at 0:12

$\begingroup$The probability that the draw is from the first distribution is either $0$ or $1$ but no one except you can tell which value is the correct one.$\endgroup$
– Dilip SarwateMar 4 '16 at 3:23

3 Answers
3

As Adrian already suggested you need to know the prior probability that $X$ came from each distribution. If $Y$ is an indicator telling us whether or not $X$ came from distribution one and $p_1$ and $p_2$ are the mixing (prior) probabilities then

Anything to the left of this line is more likely to come from the distribution with the smaller mean, while anything above this line is more likely to come from the distribution with the larger mean. In the picture you've shown if the value $s$ actually does come from the distribution with the smaller mean, and it falls below $\mu_1$ then we are almost certain it comes from $X_1 \sim N(\mu_1,\sigma^2)$ since the pdf of $X_2$ is basically 0 at this point (i.e. $f_{X_2}(s)\approx 0, s<\mu_1)$

$\begingroup$As comments and a good reply indicate, this answer implicitly assumes the two distributions are drawn with equal probability. It is crucial to make this assumption clear.$\endgroup$
– whuber♦Jul 19 '18 at 12:50

I'll try giving a very low-level answer. If there's no prior reason to believe that s was drawn from X1 or X2, i.e. the value of s is the only information you have, then the probability is directly obtainable from the relative values of the probability density functions evaluated at s. That is, $P(X_1 | s)\propto X_1(x=s) / X2(x=s)$. (The actual probability would require some normalization.)

For example, if s=0 in your example, it's much more likely that your draw was from the solid line distribution (value ~ 0.8) than the dashed line distribution (value ~ 0.1), and the relative probability is approximately 8:1. i.e. It's 8 times more likely that s=0 was drawn from the solid line distribution than the dashed line.